Amazon Interview Question
Country: United States
Interview Type: Phone Interview
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I. At first find found all primes <= N (sieve of Eratosthenes). Getting the sum will be easy then.
Follow-up:
Cache the sums for any given N to save time. {N:SUM}
Optimization: Don't have to store sums for every N.
When N = 7, N = 8, N = 9, N = 10, the prime sum remains 17.
For N between 11 to 12, the prime sum is 28.
For N between 13 to 16, the sum is 41.
Use a BST structure as the cache. For N = 16, cache:
{2:3, 4:6, 6:11, 10:17, 12:28, 16:41}
For a given N, call cache.ceilingKey(N) to find the bucket for N.
N/log(n) * log(N)
Complexity
Time:
sieve of Eratosthenes takes O(NloglogN) time.
Insert an element into BST takes O(logN), there are N/logN primes in total to be added.
So building the cache takes logN * N / LogN = O(N) time
requesting primeSum(N) takes O(logN)
Space:
sieve of Eratosthenes takes O(N) extra space which will later be release after the cache is created.
Cache: O(N/logN)
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